Year 9 Maths The Thai Guy

Lesson 5.1 · 25 min

The Mean and Range from Grouped Data

Using midpoints to estimate the mean of a grouped frequency table, and estimating the range from the class boundaries.

In this lesson you will learn to…

  • Find the midpoint of a class interval
  • Estimate the mean from a grouped frequency table using
  • Estimate the range from the first and last class boundaries
  • Explain why these values are only estimates

Why this matters

Large data sets — the heights of every Year 9 student in a country, the waiting times at a hospital — are almost always published in groups. You rarely see the raw data. Knowing how to get a good estimate of the mean from a table is how researchers and journalists make sense of those figures.

Setting out the table

Add two columns to the frequency table — one for the midpoint and one for f×xf \times x:

Height (hh, cm)Frequency, ffMidpoint, xxf×xf \times x
140≤h<145140 \le h < 1451142.5142.5
145≤h<150145 \le h < 1503147.5442.5
150≤h<155150 \le h < 15511152.51677.5
155≤h<160155 \le h < 1607157.51102.5
160≤h<165160 \le h < 1652162.5325
165≤h<170165 \le h < 1700167.50
170≤h<175170 \le h < 1751172.5172.5
Total253862.5

mean≈ΣfxΣf=3862.525=154.5 cm\text{mean} \approx \frac{\Sigma fx}{\Sigma f} = \frac{3862.5}{25} = 154.5\ \text{cm}

The Greek letter Σ\Sigma (sigma) just means “add them all up”.

A quick sanity check

The mean must lie inside the range of the data, and it is usually near the classes with the biggest frequencies. For 40 oranges with masses between 120 g and 145 g, a mean of 115 g or 145 g is impossible — the first is below every orange, and the second would need every orange to be at the very top of the last class.

Common slip. Dividing Σfx\Sigma fx by the number of classes instead of the total frequency. There are 7 classes above, but 25 children.

Key rules and formulas

  • Midpoint of a class

    For the midpoint is .

  • Estimated mean

    Multiply each midpoint by its frequency , add up the results (), then divide by the total frequency ().

  • Estimated range

    This gives the largest the range could possibly be.

  • Why only an estimate?

    Once data is grouped, the exact values are lost. Using the midpoint assumes every value in a class sits in the middle of it — which is almost never exactly true.
Try it

Grouped data explorer

04812161020304050Height, h (cm)mean ≈ 32
Total frequency, Σf
30
Σfx (midpoint × frequency)
960
Estimated mean = Σfx ÷ Σf
32
Modal class
30 ≤ x < 40
Median class (value number 15.5)
30 ≤ x < 40
Estimated range
40

These are the heights of 30 plants from the first worked example. Drag a slider to change a frequency and watch the estimated mean move towards the bigger bars.

Worked examples

Worked example 1

Estimating a mean

The heights of 30 plants are grouped: : 4, : 8, : 11, : 7. Estimate the mean height.

Try it yourself first, then reveal the solution one step at a time.

Worked example 2

Estimating the range

For the same plants, estimate the range.

Try it yourself first, then reveal the solution one step at a time.

Practice: mean and range from grouped data

9 questions

Nine questions. A calculator is fine here — the skill is setting out the table correctly.
  1. Question 1 of 9Quick-fire drill
    Quick-fire: the midpoint of each class.

    8 questions · 60 seconds · answer as many as you can.

  2. Question 2 of 9Complete the tableFoundation
    Complete the table for the heights of 30 plants.
    Height (h, cm)Frequency, fMidpoint, xf × x
    10 ≤ h < 204
    20 ≤ h < 308
    30 ≤ h < 4011
    40 ≤ h < 507
    Total30
  3. Question 3 of 9Step by step
    Estimate the mean one step at a time.
    The masses of 40 oranges: : 5, : 16, : 13, : 4, : 2.
    1. What is the midpoint of the class ?

  4. Question 4 of 9Type the answer
    Students' distances from school: : 10, : 15, : 7, : 2, : 1. Estimate the mean distance, to 2 decimal places.
    km
  5. Question 5 of 9Type the answer
    For the same students, estimate the range of distances.
    km
  6. Question 6 of 9Multiple choice
    The masses of 40 oranges all lie between 120 g and 145 g. Sarah thinks the mean is about 115 g and Freddie thinks it is over 140 g. Why are both wrong?
  7. Question 7 of 9Choose all that apply
    Why are the mean and range from a grouped table only estimates? Select all that apply.

    Select every correct answer — there may be more than one.

  8. Question 8 of 9Type the answerChallenge
    A village census: ages –: 92, –: 88, –: 85, –: 68, –: 55, –: 52, –: 42, –: 18 (each class includes its lower boundary). Estimate the mean age.
    years
  9. Question 9 of 9Explore and match
    The first three classes of the plant data are shown. The last class, , should make the total frequency 30. Set it correctly.

    Move the controls until you match the question, then press Check.

    04812161020304050Height, h (cm)mean ≈ 28
    Total frequency, Σf
    23
    Σfx (midpoint × frequency)
    645
    Estimated mean = Σfx ÷ Σf
    28.04
    Modal class
    30 ≤ x < 40
    Median class (value number 12)
    20 ≤ x < 30
    Estimated range
    40